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Linear Equations

1.1 Systems of Linear Equations

A linear equation in the variables x1,…,xnx_1,…,x_n is an equation that can be written in the form

a1x1+a2x2+...+anxn=ba_1x_1+a_2x_2+...+a_nx_n=b

A linear system is consistent if it has either one or infinitely many solutions. Otherwise, it is inconsistent if it has no solution.

Example: Find values of h such that the matrix is the augmented matrix of a consistent linear system.

A=[1h−3−246]A = \begin{bmatrix} 1 & h & -3\\ -2 & 4 & 6 \end{bmatrix}

A linear system is consistent if it has one or infinitely many solutions. Mathematically, we find a matrix to be inconsistent if it has a row where all the entries are zeros except for the b-entry. We see if this is the case by row-reducing the matrix.

R2+2R1→[1h−304+2h0]R_2 + 2R_1 \rightarrow \begin{bmatrix} 1 & h & -3\\ 0 & 4+2h & 0 \end{bmatrix}

We find that this matrix can’t possibly be inconsistent.

1.2 Row Reduction and Echelon Forms

Echelon Form (REF)

Reduced Echelon Form (RREF)

Theorem 1: Uniqueness of the Reduced Echelon Form

Each matrix is row equivalent to one and only one reduced echelon (RREF) matrix

Solutions of Linear Systems

Once a system is in RREF, you can write a general solution form as a linear combination of the free variables.

Example: Find the general solution for [1−20001∣ −4−7]\left[ \begin{matrix} 1 & -2 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \left| \, \begin{matrix} -4 \\ -7 \\ \end{matrix} \right. \right]

1x1−2x2+0x3=−40x1+0x2+1x3=−71x_1 - 2x_2 + 0x_3 = -4\\ 0x_1 + 0x_2 + 1x_3 = -7\\

We find that x2x_2 is the free variable, thus we write the general solution to the system in terms of the free variable.

[x1x2x3]=[2x2−4x2x−7]=x2[210]+[−40−7]\begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix}2x_2-4 \\ x_2 \\ x-7 \end{bmatrix} = x_2 \begin{bmatrix}2 \\ 1 \\ 0 \end{bmatrix} + \begin{bmatrix}-4\\ 0 \\ -7 \end{bmatrix}

Questions

Find the general solution for [1−2−13−6−2∣ 32]\left[ \begin{matrix} 1 & -2 & -1 \\ 3 & -6 & -2 \\ \end{matrix} \left| \, \begin{matrix} 3 \\ 2 \\ \end{matrix} \right. \right]

First, we row reduce to RREF.

R2−3R1→[1−2−1001∣ 3−7]R1+R2→[1−20001∣ −4−7]R_2 -3R1 \rightarrow \left[ \begin{matrix} 1 & -2 & -1 \\ 0 & 0 & 1 \\ \end{matrix} \left| \, \begin{matrix} 3 \\ -7 \\ \end{matrix} \right. \right]\\ R_1 + R2 \rightarrow \left[ \begin{matrix} 1 & -2 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \left| \, \begin{matrix} -4 \\ -7 \\ \end{matrix} \right. \right]

Then write as a general solution, which I’ve already done in the example above.

1.3 Vector Equations

Span

The set of all linear combinations of v1,…,vpv_1,\dots, v_p

Span(v1,…,vp)=c1v1+c2v2+⋯+cpvpSpan(v_1,\dots,v_p) = c_1v_1+c_2v_2 + \dots + c_pv_p

Questions

Determine if bb is a linear combination of a1a_1, a2a_2, a3a_3.

let a1=[1−22],a2=[055],a3=[208],b=[−511−7]\text{let } a_1 = \begin{bmatrix}1 \\ -2 \\ 2\end{bmatrix}, a_2 = \begin{bmatrix}0 \\ 5 \\ 5\end{bmatrix}, a_3 = \begin{bmatrix}2 \\ 0 \\ 8\end{bmatrix}, b = \begin{bmatrix}-5 \\ 11 \\ -7\end{bmatrix}

This is equivalent to asking if bb is in Span(a1,a2,a3)Span(a_1,a_2,a_3). Just solve for a solution xx for Ax=bAx=b where the columns of AA are the vectors aia_i.

1.4 The Matrix Equation Ax=bAx=b

Ax=[a1a2…an][x1⋮xn]=x1a1+x2a2+⋯+xnanAx = \begin{bmatrix}a_1 & a_2 & \dots & a_n\end{bmatrix} \begin{bmatrix} x_1 \\ \vdots \\ x_n\end{bmatrix} = x_1a_1 + x_2a_2 + \dots + x_na_n
A(u+v)=Au+avA(cu)=c(Au)A(u+v) = Au+av\\ A(cu) = c(Au)

The equation Ax=bAx=b has a solution if and only if bb is a linear combination of the columns of AA.

Questions

For a matrix B∈R4×4B \in R^{4\times4}, do the columns of BB span R4R^4? Does the equation Bx=yBx=y have a solution for each yy in R4R^4?

This is equivalent to asking if the columns of BB are linearly independent.

1.5 Solution Sets of Linear Systems

The homogeneous equation Ax=0Ax=0 has a nontrivial solution if and only if the equation has at least one free variable (linearly dependent).

Theorem 6: Solutions of Nonhomogeneous Systems

The solution for Ax=bAx=b is equivalent to the solution Ax=0Ax=0 transposed by some vector pp.

x=p+tvx = p+tv

Questions

Write the solution set of the following homogeneous system in parametric vector form.

x1+3x2−5x3=0x1+4x2−8x3=0−3x1−7x2+9x3=0x_1 + 3x_2 - 5x_3 = 0\\ x_1 + 4x_2 -8x_3 = 0\\ -3x_1 -7x_2 + 9x_3 = 0

Row-reduce, write the general form solution in parametric vector form:

x={t(4,−3,1)∣t∈R}x = \{t(4,-3,1) | t \in R\}

Describe the solutions of Ax=0Ax=0 in parametric vector form for A=[1−2−95012−6]A= \begin{bmatrix}1 & -2 & -9 & 5\\ 0 & 1 & 2 & -6\end{bmatrix}

Row reduce into RREF:

A=[10−5−7012−6]A= \begin{bmatrix}1 & 0 & -5 & -7\\ 0 & 1 & 2 & -6\end{bmatrix}

General solution:

x=[x1x2x3]=[5x3+7x4−2x3+6x4x3]=x3[5−21]+x4[760]x = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} = \begin{bmatrix}5x_3 + 7x_4\\ -2x_3 + 6x_4\\ x_3 \end{bmatrix} = x_3\begin{bmatrix}5 \\ -2 \\ 1\end{bmatrix} + x_4\begin{bmatrix}7 \\ 6 \\ 0\end{bmatrix}
x={s(5,−2,1)+t(7,6,0)∣s,t∈R}x = \{s(5,-2,1) + t(7,6,0) | s,t\in R\}

Find the parametric equation of the line a=[3−4]a= \begin{bmatrix}3 \\ -4\end{bmatrix} parallel to b=[−78]b= \begin{bmatrix}-7 \\ 8\end{bmatrix}

The line parallel to bb is given by bb itself, to make sure this line is through the point aa, you shift it by aa

x=a+tbx = a +tb

Find a parametric equation of the line MM through pp and qq

The line that goes runs parallel to pp and qq is given by p−qp-q. To make sure it actually goes through pp and qq, we just add either vector to it

x=p+t(p−q)x = p +t(p-q)

1.7 Linear Independence

An indexed set of vectors {v1,…,vp}∈Rn\{v_1,\dots,v_p\} \in R^n is linearly independent if the vector equation

x1v1+x2v2+⋯+xpvp=0⃗x_1v_1 + x_2v_2 + \dots + x_pv_p = \vec 0

has only the trivial solution x=0⃗x = \vec 0.

A set of two vectors {v1,v2}\{v_1, v_2\} is linearly dependent if one of the vectors is a multiple of the other.

Theorem 7: Characterization of Linearly Dependent Sets

An indexed set S={v1,…,vp}∈RnS=\{v_1,\dots,v_p\} \in R^n is linearly dependent iff at least one of the vectors in SS is a linear combination of the others.

Theorem 8

If a set S={v1,…,vp}∈RnS=\{v_1,\dots,v_p\} \in R^n contains more vectors than there are entries in each vector (p>np> n), then the set is linearly dependent. Matrices that are horizontally rectangular have linearly dependent column vectors.

Theorem 9

If a set S={v1,…,vp}∈RnS=\{v_1,\dots,v_p\} \in R^n contains the zero vector, then the set is linearly dependent because the equation x1v1+x2v2+⋯+xpvp=0⃗x_1v_1 + x_2v_2 + \dots + x_pv_p = \vec 0 has a non-trivial solution.

Questions

True or False: The columns of A∈R4×5A\in\R^{4\times5} are linearly dependent.

True by Theorem 8.

True or False: If xx and yy are linearly independent, and {x,y,z}\{x,y,z\} is linearly dependent, then zz is in Span{x,y}Span\{x,y\}

True. There exists some combination c1x+c2y=c3zc_1x+c_2y=c_3z

5, 9, 11, 17, 25, 27.

1.8 Introduction to Linear Transformations

A transformation (or function or mapping) TT from RnR^n to RmR^m is a rule that assigns to each vector xx in RnR^n a vector T(x)T(x) in RmR^m.

Matrix Transformations

T(x):x↦AxA∈Rm×nT(x): x \mapsto Ax\\ A \in R^{m\times n}

Example: Let A∈R3×2A \in R^{3\times2}, u∈R2u\in R^2, b,c∈R3b ,c \in R^3.

T(u)=AuT(u) = Au

Linear Transformations

Given that a linear transformation is defined by a matrix transformation x↦Axx \mapsto Ax, it has the following properties:

T(u+v)=T(u)+T(v)T(cu)=cT(u)T(u+v) = T(u) + T(v)\\ T(cu) = cT(u)

Additionally:

T(0)=0T(cu+dv)=cT(u)+T(v)T(0) = 0\\ T(cu+dv) = cT(u) + T(v)

Onto

Rank(A)=m\text{Rank}(A) = m

One-to-One

Rank(A)=n\text{Rank}(A) = n

Questions

Let A∈R4×3A\in R^{4\times 3}, b∈R4b\in R^4, T(x)=AxT(x) = Ax. Find a vector xx whose image under TT is bb.

Just solve Ax=bAx=b

Describe geometrically what TT does to each vector xx in R2R^2. T(x)=[0110][x1x2]T(x) = \begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix}x_1\\x_2 \end{bmatrix}.

It flips the xx and yy values, resulting in a reflection along [11]\begin{bmatrix}1 \\ 1\end{bmatrix}

True or False: A linear transformation preserves the operations of vector addition and scalar multiplication

True by definition.

True or False: Every matrix transformation is a linear transformation

True. Ax is closed under vector addition and scalar multiplication.

1.9 The Matrix of a Linear Transformation

Theorem 10

A linear transformation T(X)T(X) is defined by a unique matrix AA such that

A=[T(e1)…T(en)]A = \begin{bmatrix}T(e_1) \dots T(e_n)\end{bmatrix}

Onto

If each b∈Rmb \in \R^m is the image of at least one x∈Rnx \in \R^n, the mapping is onto. Ax=bAx=b has at least one solution for each bb (spans the codomain) if and only if RankA=m\text{Rank}A = m

One to One

If each b∈Rmb \in \R^m is the image of at most one x∈Rnx \in \R^n, the mapping is one to one. Ax=bAx=b has exactly one solution for each bb.

Theorem 11

TT is one to one if and only if the T(x)=0T(x)=0 has only the trivial solution.

Theorem 12

Example: Is the linear transformation defined by A onto or one-to-one?

A=[1−48102−130005]A = \begin{bmatrix} 1 & -4 & 8 & 1\\ 0 & 2 & -1 & 3\\ 0 & 0 & 0 & 5 \end{bmatrix}

The matrix is full rank, thus it spans the codomain, so we know that it is onto.

By Theorem 11 we know that TT is only one-to-one if the columns of AA are linearly independent. By Theorem 8, we know that when there are more columns than rows, the columns are automatically linearly dependent, so it’s not one-to-one.